Problems

Problem :
Find the critical points and inflection points of the function
f (x) = x4 -2x2
(with domain
the set of all real numbers). Which of the critical points are local minima? local
maxima? Is there a global minimum or maximum?

We first calculate the derivatives of the function:

f'(x)

=

4x3 - 4x

=

4(x + 1)x(x - 1)

f''(x)

=

12x2 - 4

=

4(3x2 - 1)

We see that
f'(x) = 0
when
x = - 1
,
0
, or
1
, so these are the three critical points of
f
. We calculate the second derivatives at these points:

f''(- 1)

=

8

f''(0)

=

-4

f''(1)

=

8

so by the second derivative test,
f
has local minima at
-1
and
1
and a local maximum
at
0
. Substituting back into the original function yields

f (- 1)

=

-1

f (0)

=

0

f (1)

=

-1

so
f
attains its global minimum of
-1
at
x = ±1
. It is clear from the graph of
f
that it has no global maximum.

Figure %: Graph of
f (x) = x4 -2x2

To find the points of inflection, we solve
f''(x) = 0
, or
12x2 - 4 = 0
, which has
solutions
x = ±1/3) ±0.58
. Referring once again to the graph of
f
,
we can check that the concavity does indeed change at these
x
-values.

Problem :
Find the inflection points of
f (x) = e-x2
. (This famous function
is called the gaussian.)

We compute the derivatives:

f'(x)

=

-2xe-x2,

f''(x)

=

(- 2x)(- 2xe-x2) + (- 2)(e-x2)

=

(4x2 -2)e-x2

Solving
f''(x) = 0
for
x
, we get the inflection points
x = ±1/ 0.71
.
This is reasonable when one considers the graph of
f
.